1. The solutions to \(\cos (u) =-\frac {\sqrt {3}}{2}\) are \(u = \frac {5\pi }{6} + 2\pi k\) or \(u = \frac {7\pi }{6} + 2\pi k\) for integers \(k\).

   Since the argument of cosine here is \(2\theta \), this means \(2\theta = \frac {5\pi }{6} + 2\pi k\) or \(2\theta = \frac {7\pi }{6} + 2\pi k\) for integers \(k\). Solving for \(\theta \) gives \(\theta = \frac {5\pi }{12} + \pi k\) or \(\theta = \frac {7\pi }{12} + \pi k\) for integers \(k\).

   To check these answers analytically, we substitute them into the original equation. For any integer \(k\):

   \[ \begin{array}{rclr} \cos \left ( 2\left [\frac {5\pi }{12} + \pi k\right ]\right ) & = & \cos \left (\frac {5\pi }{6} + 2\pi k\right ) & \\ & = & \cos \left (\frac {5\pi }{6}\right ) & \text {(the period of cosine is $2\pi $)} \\ & = & -\frac {\sqrt {3}}{2} & \\ \end{array}\]

   Similarly, we find \(\cos \left ( 2\left [\frac {7\pi }{12} + \pi k\right ]\right ) = \cos \left (\frac {7\pi }{6} + 2\pi k\right ) = \cos \left (\frac {7\pi }{6}\right ) = -\frac {\sqrt {3}}{2}\).

   To determine which of our solutions lie in \([0,2\pi )\), we substitute integer values for \(k\). The solutions we keep come from the values of \(k = 0\) and \(k =1\) and are \(\theta = \frac {5\pi }{12}\), \(\frac {7\pi }{12}\), \(\frac {17\pi }{12}\) and \(\frac {19\pi }{12}\).

   Using desmos, we graph \(y = \cos (2\theta )\) and \(y = -\frac {\sqrt {3}}{2}\) over \([0,2\pi )\) and examine where these two graphs intersect to verify our answers.

2. Since this equation has the form \(\csc (u) = \sqrt {2}\), we rewrite this as \(\sin (u) = \frac {\sqrt {2}}{2}\) and find \(u = \frac {\pi }{4} + 2\pi k\) or \(u = \frac {3\pi }{4} + 2\pi k\) for integers \(k\).

   Since the argument of cosecant here is \(\left (\frac {1}{3}\theta -\pi \right )\), \(\frac {1}{3}\theta -\pi = \frac {\pi }{4} + 2\pi k\) or \(\frac {1}{3}\theta - \pi = \frac {3\pi }{4} + 2\pi k\).

   To solve \(\frac {1}{3} \theta -\pi = \frac {\pi }{4} + 2\pi k\), we first add \(\pi \) to both sides to get \(\frac {1}{3} \theta = \frac {\pi }{4} + 2\pi k + \pi \). A common error is to treat the ‘\(2\pi k\)’ and ‘\(\pi \)’ terms as ‘like’ terms and try to combine them when they are not.

   We can, however, combine the ‘\(\pi \)’ and ‘\(\frac {\pi }{4}\)’ terms to get \(\frac {1}{3} \theta = \frac {5\pi }{4} + 2\pi k\).

   We now finish by multiplying both sides by \(3\) to get \(\theta = 3 \left ( \frac {5\pi }{4} + 2\pi k \right ) = \frac {15 \pi }{4} + 6\pi k\), where \(k\), as always, runs through the integers.

   Solving the other equation, \(\frac {1}{3} \theta -\pi = \frac {3\pi }{4} + 2\pi k\) produces \(\theta = \frac {21\pi }{4} + 6 \pi k\) for integers \(k\). To check the first family of answers, we substitute, combine line terms, and simplify.

   \[ \begin{array}{rclr} \csc \left (\frac {1}{3} \left [ \frac {15\pi }{4} + 6 \pi k \right ] - \pi \right ) & = & \csc \left (\frac {5\pi }{4} + 2\pi k - \pi \right ) & \\ & = & \csc \left (\frac {\pi }{4} + 2\pi k\right ) & \\ & = & \csc \left (\frac {\pi }{4}\right ) & \text {(the period of cosecant is $2\pi $)} \\ & = & \sqrt {2} & \\ \end{array}\]

   The family \(\theta = \frac {21\pi }{4} + 6 \pi k\) checks similarly.

   Despite having infinitely many solutions, we find that none of them lie in \([0,2\pi )\).

   To verify this graphically, we check that \(y = \csc \left (\frac {1}{3} \theta -\pi \right )\) and \(y =\sqrt {2}\) do not intersect at all over the interval \([0,2\pi )\).

3. Since \(\cot (3t) = 0\) has the form \(\cot (u) = 0\), we know \(u = \frac {\pi }{2} + \pi k\), so, in this case, \(3t = \frac {\pi }{2} + \pi k\) for integers \(k\).

   Solving for \(t\) yields \(t = \frac {\pi }{6} + \frac {\pi }{3} k\). Checking our answers, we get

   \[ \begin{array}{rclr} \cot \left (3\left [ \frac {\pi }{6} + \frac {\pi }{3} k\right ]\right ) & = & \cot \left (\frac {\pi }{2} + \pi k\right ) & \\ & = & \cot \left (\frac {\pi }{2}\right ) & \text {(the period of cotangent is $\pi $)} \\ & = & 0 & \\ \end{array}\]

   As \(k\) runs through the integers, we obtain six answers, corresponding to \(k=0\) through \(k=5\), which lie in \([0, 2\pi )\): \(x = \frac {\pi }{6}\), \(\frac {\pi }{2}\), \(\frac {5\pi }{6}\), \(\frac {7\pi }{6}\) , \(\frac {3\pi }{2}\) and \(\frac {11\pi }{6}\).

   Graphing \(y = \cot (3t)\) and \(y=0\) (the \(t\)-axis), we confirm our result. ^[2] (On many calculators, there is no function button for cotangent. In that case, we would use the quotient identity and graph \(y = \frac {\cos (3t)}{\sin (3t)}\) instead. The reader is invited to see what happens if we would graph \(y= \frac {1}{\tan (3t)}\) instead.)

4. The complication in solving an equation like \(\sec ^{2}(t) = 4\) comes not from the argument of secant, which is just \(t\), but rather, the fact the secant is being squared: \(\sec ^{2}(t) = (\sec (t))^2 = 4\).

   To get this equation to look like one of the forms listed on page 1, we extract square roots to get \(\sec (t) = \pm 2\). Converting to cosines, we have \(\cos (t) = \pm \frac {1}{2}\).

   For \(\cos (t) = \frac {1}{2}\), we get \(t = \frac {\pi }{3} + 2\pi k\) or \(t = \frac {5\pi }{3} + 2\pi k\) for integers \(k\). For \(\cos (t) = -\frac {1}{2}\), we get \(t = \frac {2\pi }{3} + 2\pi k\) or \(t = \frac {4\pi }{3} + 2\pi k\) for integers \(k\).

   If we take a step back and think of these families of solutions geometrically, we see we are finding the measures of all angles with a reference angle of \(\frac {\pi }{3}\). As a result, these solutions can be combined and we may write our solutions as \(t = \frac {\pi }{3} + \pi k\) and \(t = \frac {2\pi }{3} + \pi k\) for integers \(k\).

   To check the first family of solutions, we note that, depending on the integer \(k\), \(\sec \left (\frac {\pi }{3} + \pi k\right )\) doesn’t always equal \(\sec \left (\frac {\pi }{3}\right )\). It is true, though, that for all integers \(k\), \(\sec \left (\frac {\pi }{3} + \pi k\right ) = \pm \sec \left (\frac {\pi }{3}\right ) = \pm 2\). (Can you show this?) Hence, checking our first family of solutions gives:

   \[ \begin{array}{rclr} \sec ^{2}\left (\frac {\pi }{3} + \pi k\right ) & = & \left ( \pm \sec \left (\frac {\pi }{3}\right )\right )^2 & \\ & = & (\pm 2)^2 & \\ & = & 4 & \\ \end{array}\]

   The check for the family of solutions \(t =\frac {2\pi }{3} + \pi k\) is similar.

   The solutions which lie in \([0,2\pi )\) come from the values \(k = 0\) and \(k=1\), namely \(t = \frac {\pi }{3}\), \(\frac {2\pi }{3}\), \(\frac {4\pi }{3}\) and \(\frac {5\pi }{3}\). Graphing \(y = (\sec (t))^2\) and \(y=4\) confirms our results.

5. The equation \(\tan \left (\frac {x}{2}\right ) = -3\) has the form \(\tan (u) = -3\), whose solution is \(u = \arctan (-3) + \pi k\).

   Hence, \(\frac {x}{2} = \arctan (-3) + \pi k\), so \(x = 2\arctan (-3) + 2\pi k\) for integers \(k\). To check, we note

   \[ \begin{array}{rclr} \tan \left (\frac {2\arctan (-3) + 2\pi k}{2}\right ) & = & \tan \left ( \arctan (-3) + \pi k \right ) & \\ & = & \tan \left (\arctan (-3) \right ) & \text {(the period of tangent is $\pi $)} \\ & = & -3 & \\ \end{array}\]

   (See Theorem arctangentcotangentfunctionprops)

   To determine which of our answers lie in the interval \([0,2\pi )\), we first need to get an idea of the value of \(2\arctan (-3)\). While we could easily find an approximation using a calculator, ^[3] (Your instructor will let you know if you should abandon the analytic route at this point and use your calculator.) we proceed analytically, as is our custom.

   To get started, we note that since \(-3 < 0\), it \(-\frac {\pi }{2} < \arctan (-3) < 0\). Hence, \(-\pi < 2\arctan (-3) < 0\). With regard to our solutions, \(x = 2\arctan (-3) + 2\pi k\), we see for \(k = 0\), we get \(x = 2\arctan (-3) < 0\), so we discard this answer and all answers \(x = 2\arctan (-3) + 2\pi k\) where \(k < 0\).

   Next, we turn our attention to \(k = 1\) and get \(x = 2\arctan (-3) + 2\pi \). Starting with the inequality \(-\pi < 2\arctan (-3) < 0\), we add through \(2\pi \) and get \(\pi < 2\arctan (-3) +2\pi < 2\pi \). This means \(x = 2\arctan (-3) + 2\pi \) lies in \([0,2\pi )\).

   Advancing \(k\) to \(2\) produces \(x = 2\arctan (-3) + 4\pi \). Once again, we get from \(-\pi < 2\arctan (-3) < 0\) that \(3\pi < 2\arctan (-3) + 4\pi < 4\pi \). Since this is outside the interval of interest, \([0,2\pi )\), we discard \(x = 2\arctan (-3) + 4\pi \) and all solutions of the form \(x = 2\arctan (-3) + 2\pi k\) for \(k > 2\).

   Graphically, \(y = \tan \left (\frac {x}{2}\right )\) and \(y = -3\) intersect only once on \([0,2\pi )\) at \(x = 2\arctan (-3) + 2\pi \approx 3.785\).

6. To solve \(\sin (2x) = 0.87\), we first note that it has the form \(\sin (u) = 0.87\), which has the family of solutions \(u = \arcsin (0.87) + 2\pi k\) or \(u =\pi - \arcsin (0.87) + 2\pi k\) for integers \(k\).

   Since the argument of sine here is \(2x\), we get \(2x = \arcsin (0.87) + 2\pi k\) or \(2x =\pi - \arcsin (0.87) + 2\pi k\) which gives \(x = \frac {1}{2} \arcsin (0.87) + \pi k\) or \(x =\frac {\pi }{2} - \frac {1}{2}\arcsin (0.87) + \pi k\) for integers \(k\). To check,

   \[ \begin{array}{rclr} \sin \left (2\left [\frac {1}{2} \arcsin (0.87) + \pi k\right ]\right ) & = & \sin \left (\arcsin (0.87) + 2\pi k\right ) & \\ & = & \sin \left (\arcsin (0.87)\right ) & \text {(the period of sine is $2\pi $)} \\ & = & 0.87& (\text {See Theorem } \ref {arccosinesinefunctionprops})\\ \end{array}\]

   For the family \(x =\frac {\pi }{2} - \frac {1}{2}\arcsin (0.87) + \pi k\) , we get

   \[ \begin{array}{rclr} \sin \left (2\left [\frac {\pi }{2} - \frac {1}{2} \arcsin (0.87) + \pi k\right ]\right ) & = & \sin \left (\pi - \arcsin (0.87) + 2\pi k\right ) & \\ & = & \sin \left (\pi - \arcsin (0.87)\right ) & \text {(the period of sine is $2\pi $)} \\ & = & \sin \left (\arcsin (0.87)\right ) & \text {($\sin (\pi - t) = \sin (t)$)} \\ & = & 0.87& (\text {See Theorem } \ref {arccosinesinefunctionprops}) \\ \end{array}\]

   To determine which of these solutions lie in \([0,2\pi )\), we first need to get an idea of the value of \(x=\frac {1}{2} \arcsin (0.87)\). Once again, we could use the calculator, but we adopt an analytic route here.

   By definition, \(0 < \arcsin (0.87) < \frac {\pi }{2}\) so that multiplying through by \(\frac {1}{2}\) gives us \(0 < \frac {1}{2} \arcsin (0.87) < \frac {\pi }{4}\).

   Starting with the family of solutions \(x = \frac {1}{2} \arcsin (0.87) + \pi k\), we use the same kind of arguments as in our solution to number arctanin02pi above and find only the solutions corresponding to \(k =0\) and \(k=1\) lie in \([0,2\pi )\): \(x = \frac {1}{2} \arcsin (0.87)\) and \(x = \frac {1}{2} \arcsin (0.87) + \pi \).

   Next, we move to the family \(x =\frac {\pi }{2} - \frac {1}{2}\arcsin (0.87) + \pi k\) for integers \(k\). Here, we need to get a better estimate of \(\frac {\pi }{2} - \frac {1}{2} \arcsin (0.87)\). From the inequality \(0 < \frac {1}{2}\arcsin (0.87) < \frac {\pi }{4}\), we first multiply through by \(-1\) and then add \(\frac {\pi }{2}\) to get \(\frac {\pi }{2} > \frac {\pi }{2} -\frac {1}{2} \arcsin (0.87) > \frac {\pi }{4}\), or \(\frac {\pi }{4} < \frac {\pi }{2} -\frac {1}{2} \arcsin (0.87) < \frac {\pi }{2}\).

   Proceeding with the usual arguments, we find the only solutions which lie in \([0,2\pi )\) correspond to \(k = 0\) and \(k=1\), namely \(x =\frac {\pi }{2} - \frac {1}{2}\arcsin (0.87)\) and \(x = \frac {3\pi }{2} - \frac {1}{2}\arcsin (0.87)\).

   All told, we have found four solutions to \(\sin (2x) = 0.87\) in \([0,2\pi )\): \(x =\frac {1}{2} \arcsin (0.87) \approx 0.528\), \(x=\frac {1}{2} \arcsin (0.87) + \pi \approx 3.669\), \(x =\frac {\pi }{2} - \frac {1}{2}\arcsin (0.87) \approx 1.043\) and \(x = \frac {3\pi }{2} - \frac {1}{2}\arcsin (0.87) \approx 4.185\). By graphing \(y = \sin (2x)\) and \(y = 0.87\), we confirm our results.

1. One approach to solving \(3\sin ^{3}(\theta ) = \sin ^{2}(\theta )\) begins with dividing both sides by \(\sin ^{2}(\theta )\). Doing so, however, assumes that \(\sin ^{2}(\theta ) \neq 0\) which means we risk losing solutions.

   Instead, we take a cue from Chapter PolynomialFunctions (since what we have here is a polynomial equation in terms of \(\sin (\theta )\)) and gather all the nonzero terms on one side and factor:

   \[ \begin{array}{rclr} 3\sin ^{3}(\theta ) & = & \sin ^{2}(\theta ) & \\ 3\sin ^{3}(\theta ) - \sin ^{2}(\theta ) & = & 0 & \\ \sin ^{2}(\theta ) (3 \sin (\theta ) - 1) & = & 0 & \text {Factor out $\sin ^{2}(\theta )$ from both terms.} \\ \end{array} \]

   We get \(\sin ^{2}(\theta ) = 0\) or \(3\sin (\theta ) - 1 = 0\), so \(\sin (\theta ) = 0\) or \(\sin (\theta ) = \frac {1}{3}\). The solution to \(\sin (\theta ) = 0\) is \(\theta = \pi k\), with \(\theta = 0\) and \(\theta = \pi \) being the two solutions which lie in \([0,2\pi )\).

   To solve \(\sin (\theta ) = \frac {1}{3}\), we use the arcsine function to get \(\theta = \arcsin \left (\frac {1}{3}\right ) + 2\pi k\) or \(\theta = \pi - \arcsin \left (\frac {1}{3}\right ) + 2\pi k\) for integers \(k\). We find the two solutions here which lie in \([0,2\pi )\) to be \(\theta = \arcsin \left (\frac {1}{3}\right ) \approx 0.34\) and \(\theta = \pi - \arcsin \left (\frac {1}{3}\right ) \approx 2.80\).

   To check graphically, we plot \(y = 3(\sin (\theta ))^3\) and \(y = (\sin (\theta ))^2\) and find the \(\theta \)-coordinates of the intersection points of these two curves. ^[4] (Note that we do not list \(\theta = 2\pi \) as part of the solution over the interval \([0,2\pi )\) since \(2\pi \) is not in \([0, 2\pi )\).) (Some extra zooming may be required near \(\theta =0\) and \(\theta =\pi \) to verify that these two curves do in fact intersect four times.)

2. We see immediately in the equation \(\sec ^{2}(\theta ) = \tan (\theta ) + 3\) that there are two different circular functions present, so we look for an identity to express both sides in terms of the same function.

   We use the Pythagorean Identity \(\sec ^{2}(\theta ) = 1 + \tan ^{2}(\theta )\) to exchange \(\sec ^{2}(\theta )\) for tangents. What results is a ‘quadratic in disguise:’ ^[5] (See Section AppQuadEqus for a review of this concept.)

   \[ \begin{array}{rclr} \sec ^{2}(\theta ) & = & \tan (\theta ) + 3 & \\ 1 + \tan ^{2}(\theta ) & = & \tan (\theta ) + 3& \text {(Since $\sec ^{2}(\theta ) = 1 + \tan ^{2}(\theta )$.)} \\ \tan ^{2}(\theta ) - \tan (\theta ) -2 & = & 0 & \\ u^2 - u - 2 & = & 0 & \text {Let $u = \tan (\theta )$.} \\ (u + 1)(u - 2) & = & 0 & \\ \end{array} \]

   This gives \(u = -1\) or \(u = 2\). Since \(u = \tan (\theta )\), we have \(\tan (\theta ) = -1\) or \(\tan (\theta ) = 2\).

   From \(\tan (\theta ) = -1\), we get \(\theta = -\frac {\pi }{4} + \pi k\) for integers \(k\). To solve \(\tan (\theta ) = 2\), we employ the arctangent function and get \(\theta = \arctan (2) + \pi k\) for integers \(k\).

   From the first set of solutions, we get \(\theta = \frac {3\pi }{4}\) and \(\theta = \frac {7\pi }{4}\) as our answers which lie in \([0,2\pi )\).

   Using the same sort of argument we saw in Example TrigEqnEx1, we get \(\theta =\arctan (2) \approx 1.107\) and \(\theta = \pi + \arctan (2) \approx 4.249\) as answers from our second set of solutions which lie in \([0,2\pi )\).

   We verify our solutions below graphically.

3. The good news is that in the equation \(\cos (2t) = 3\cos (t) - 2\), we have the same circular function, cosine, throughout. The bad news is that we have different arguments, \(2t\) and \(t\).

   Using the double angle identity \(\cos (2t) = 2\cos ^{2}(t) - 1\) results in another quadratic in disguise:’

   \[ \begin{array}{rclr} \cos (2t) & = & 3\cos (t) - 2 & \\ 2\cos ^{2}(t) -1 & = & 3\cos (t) -2 & \text {(Since $\cos (2t) = 2\cos ^{2}(t) -1$.)} \\ 2\cos ^{2}(t) - 3\cos (t) +1 & = & 0 & \\ 2 u^2 - 3 u + 1 & = & 0 & \text {Let $u = \cos (t)$.}\\ (2u - 1)(u - 1) & = & 0 & \\ \end{array} \]

   We get \(u = \frac {1}{2}\) or \(u = 1\), so \(\cos (t) = \frac {1}{2}\) or \(\cos (t) = 1\). Solving \(\cos (t) = \frac {1}{2}\), we get \(t = \frac {\pi }{3} + 2\pi k\) or \(t = \frac {5\pi }{3} + 2\pi k\) for integers \(k\). From \(\cos (t) = 1\), we get \(t = 2\pi k\) for integers \(k\).

   The answers which lie in \([0,2\pi )\) are \(t =0\), \(\frac {\pi }{3}\), and \(\frac {5\pi }{3}\). Graphing \(y = \cos (2t)\) and \(y = 3\cos (t) - 2\), we find that the curves intersect in three places on \([0,2\pi )\) and confirm our results.

4. To solve \(\cos (3t) = 2- \cos (t)\), we take a cue from the previous problem and look for an identity to rewrite \(\cos (3t)\) in terms of \(\cos (t)\).

   From Example doubleangleex, number cosinepolynomial, we know that \(\cos (3t) = 4\cos ^{3}(t) - 3\cos (t)\). This transforms the equation into a polynomial in terms of \(\cos (t)\).

   \[ \begin{array}{rclr} \cos (3t) & = &2- \cos (t) & \\ 4\cos ^{3}(t) - 3\cos (t) & = & 2- \cos (t) & \\ 2\cos ^{3}(t) - 2\cos (t) -2 & = & 0 & \\ 4 u^3 - 2 u -2 & = & 0 & \text {Let $u = \cos (t)$.} \\ \end{array} \]

   Using what we know from Chapter PolynomialFunctions, we factor \(4u^3-2u-2\) as \((u-1)\left (4u^2+4u+2\right )\) and set each factor equal to \(0\).

   We get either \(u-1 = 0\) or \(4u^2+2u+2=0\), and since the discriminant of the latter is negative, the only real solution to \(4u^3-2u-2=0\) is \(u = 1\).

   Since \(u = \cos (t)\), we get \(\cos (t) = 1\), so \(t = 2\pi k\) for integers \(k\). The only solution which lies in \([0,2\pi )\) is \(t = 0\). Our graph below confirms this.

5. While we could approach solving the equation \(\cos (3x) = \cos (5x)\) in the same manner as we did the previous two problems, we choose instead to showcase the utility of the Sum to Product Identities. ^[6] (We invite the reader to try the ‘polynomial approach’ used in the previous problem to see what difficulties are encountered.)

   From \(\cos (3x) = \cos (5x)\), we get \(\cos (5x) - \cos (3x) = 0\), and it is the presence of \(0\) on the right hand side that indicates a switch to a product would be a good move. ^[7] (Since a product equalling zero means, necessarily, one or both factors is \(0\). See page propertiesofzero.)

   Using Theorem sumtoproduct, we rewrite \(\cos (5x) - \cos (3x)\) as \(- 2 \sin \left ( \frac {5x + 3x}{2}\right )\sin \left ( \frac {5x - 3x}{2}\right ) = -2 \sin (4x)\sin (x)\). Hence, our original equation \(\cos (3x) = \cos (5x)\) is equivalent to \(-2 \sin (4x) \sin (x) = 0\).

   From \(-2 \sin (4x) \sin (x) = 0\), we get either \(\sin (4x) = 0\) or \(\sin (x)\) = 0. Solving \(\sin (4x) = 0\) gives \(x = \frac {\pi }{4} k\) for integers \(k\), and the solution to \(\sin (x) = 0\) is \(x = \pi k\) for integers \(k\).

   The second set of solutions is contained in the first set of solutions, ^[8] (As always, when in doubt, write it out!) so our final solution to \(\cos (5x) = \cos (3x)\) is \(x = \frac {\pi }{4} k\) for integers \(k\).

   There are eight of these answers which lie in \([0,2\pi )\): \(x = 0\), \(\frac {\pi }{4}\), \(\frac {\pi }{2}\), \(\frac {3\pi }{4}\), \(\pi \), \(\frac {5\pi }{4}\), \(\frac {3\pi }{2}\) and \(\frac {7\pi }{4}\). Our plot of the graphs of \(y = \cos (3x)\) and \(y = \cos (5x)\) below bears this out.

6. In the equation \(\sin (2x) =\sqrt {3} \cos (x)\), we not only have different circular functions involved, but we also have different arguments to contend with.

   Using the double angle identity \(\sin (2x) = 2 \sin (x) \cos (x)\) makes all of the arguments the same and we proceed to gather all of the nonzero terms on one side of the equation and factor.

   \[ \begin{array}{rclr} \sin (2x) & = & \sqrt {3} \cos (x) & \\ 2 \sin (x) \cos (x) & = & \sqrt {3} \cos (x) & \text {(Since $\sin (2x) = 2\sin (x) \cos (x)$.)} \\ 2\sin (x) \cos (x) - \sqrt {3} \cos (x) & = & 0 & \\ \cos (x) (2 \sin (x) - \sqrt {3}) & = & 0 & \\ \end{array} \]

   We get \(\cos (x) = 0\) or \(\sin (x) = \frac {\sqrt {3}}{2}\). From \(\cos (x) = 0\), we obtain \(x = \frac {\pi }{2} + \pi k\) for integers \(k\). From \(\sin (x) = \frac {\sqrt {3}}{2}\), we get \(x = \frac {\pi }{3} + 2\pi k\) or \(x = \frac {2\pi }{3} + 2\pi k\) for integers \(k\).

   The answers which lie in \([0,2\pi )\) are \(x = \frac {\pi }{2}\), \(\frac {3\pi }{2}\), \(\frac {\pi }{3}\) and \(\frac {2\pi }{3}\), as verified graphically below.

7. Unlike the previous problem, there seems to be no quick way to get the circular functions or their arguments to match in the equation \(\sin (x)\cos \left (\frac {x}{2}\right ) + \cos (x)\sin \left (\frac {x}{2}\right ) = 1\).

   If we stare at it long enough, however, we realize that the left hand side is the expanded form of the sum formula for \(\sin \left (x + \frac {x}{2}\right )\). Hence, our original equation is equivalent to \(\sin \left (\frac {3}{2} x\right ) = 1\).

   Solving, we find \(x = \frac {\pi }{3} + \frac {4\pi }{3} k\) for integers \(k\). Two of these solutions lie in \([0,2\pi )\): \(x = \frac {\pi }{3}\) and \(x = \frac {5\pi }{3}\). Graphing \(y = \sin (x)\cos \left (\frac {x}{2}\right ) + \cos (x)\sin \left (\frac {x}{2}\right )\) and \(y = 1\) validates our solutions.

8. With the absence of double angles or squares, there doesn’t seem to be much we can do with the equation \(\cos (x) - \sqrt {3} \sin (x) = 2\).

   However, since the frequencies of the sine and cosine terms are the same, we can rewrite the left hand side of this equation as a sinusoid.

   To fit \(f(x) = \cos (x) - \sqrt {3} \sin (x)\) to the form \(A\sin (\omega t + \phi ) + B\), we use what we learned in Example expandedsinusoidex1 and find \(A = 2\), \(B = 0\), \(\omega = 1\) and \(\phi = \frac {5\pi }{6}\).

   Hence, we can rewrite the equation \(\cos (x) - \sqrt {3} \sin (x) = 2\) as \(2 \sin \left (x + \frac {5\pi }{6}\right ) = 2\), or \(\sin \left (x + \frac {5\pi }{6}\right ) = 1\). Solving, we get \(x = - \frac {\pi }{3} + 2\pi k\) for integers \(k\).

   Only one of our solutions, \(x = \frac {5\pi }{3}\), which corresponds to \(k=1\), lies in \([0,2\pi )\). Geometrically, we see that \(y = \cos (x) - \sqrt {3} \sin (x)\) and \(y = 2\) intersect just once, supporting our answer.

   An alternative way to solve this problem is to introduce squares in order to exchange sines and cosines using a Pythagorean Identity.

   From \(\cos (x) - \sqrt {3} \sin (x) = 2\) we get \(\sqrt {3} \sin (x) = \cos (x) - 2\) so that \(\left (\sqrt {3} \sin (x)\right )^2 = \left (\cos (x) - 2\right )^2\). Simplifying, we get: \(3 \sin ^{2}(x) = \cos ^{2}(x) - 4 \cos (x) + 4\).

   Substituting \(\sin ^{2}(x) = 1 - \cos ^{2}(x)\), we get \(3(1 - \cos ^{2}(x)) = \cos ^{2}(x) - 4\cos (x) + 4\) which results in the quadratic equation: \(4 \cos ^{2}(x) - 4 \cos (x) +1 = 0\).

   Letting \(u = \cos (x)\), we get \(4u^2 - 4u + 1 = 0\) or \((2u - 1)^2 = 0\). We get \(u = \cos (x) = \frac {1}{2}\). Solving \(\cos (x) = \frac {1}{2}\) gives \(x = \frac {\pi }{3} + 2\pi k\) as well as \(x = \frac {5\pi }{3} + 2\pi k\) for integers, \(k\).

   Of these two families, only solutions of the form \(x = \frac {5\pi }{3} + 2\pi k\) checks in our original equation. ^[9] (We’ve seen how squaring both sides can lead to extraneous solutions in Section AppRadEqus and Chapter RootRadicalPowerFunctions. Here, squaring both sides admits an entire family of extraneous solutions.) We leave it the reader to verify this representation of solutions to \(\cos (x) - \sqrt {3} \sin (x) = 2\) is equivalent to the one we found previously.

1. We begin solving \(2\sin (t) \leq 1\) by collecting all of the terms on one side of the equation and zero on the other to get \(2\sin (t) - 1 \leq 0\).

   Next, we let \(f(t) = 2\sin (t) - 1\) and note that our original inequality is equivalent to solving \(f(t) \leq 0\). We now look to see where, if ever, \(f\) is undefined and where \(f(t) = 0\).

   Since the domain of \(f\) is all real numbers, we can immediately set about finding the zeros of \(f\). Solving \(f(t) = 0\), we have \(2\sin (t) - 1=0\) or \(\sin (t) = \frac {1}{2}\). The solutions here are \(t = \frac {\pi }{6} + 2\pi k\) and \(t = \frac {5\pi }{6} + 2\pi k\) for integers \(k\). Since we are restricting our attention to \([0,2\pi )\), only \(t = \frac {\pi }{6}\) and \(t = \frac {5\pi }{6}\) are of concern.

   Next, we choose test values in \([0,2\pi )\) other than the zeros and determine if \(f\) is positive or negative there. For \(t = 0\) we have \(f(0) = -1\), for \(t = \frac {\pi }{2}\) we get \(f\left (\frac {\pi }{2}\right ) = 1\) and for \(t = \pi \) we get \(f(\pi ) = -1\). Our sign diagram is below.

   

   Since our original inequality is equivalent to \(f(t) \leq 0\), we are looking for where the function is negative \((-)\) or \(0\), and we get the intervals \(\left [0, \frac {\pi }{6}\right ] \cup \left [\frac {5\pi }{6}, 2\pi \right )\). We can confirm our answer graphically by seeing where the graph of \(y = 2\sin (t)\) crosses or is below the graph of \(y = 1\).

2. We first rewrite \(\sin (2x) > \cos (x)\) as \(\sin (2x) - \cos (x) > 0\) and let \(f(x) = \sin (2x) - \cos (x)\).

   Our original inequality is thus equivalent to \(f(x) > 0\). The domain of \(f\) is all real numbers, so we can advance to finding the zeros of \(f\).

   Setting \(f(x) = 0\) yields \(\sin (2x) - \cos (x) = 0\), which, by way of the double angle identity for sine, becomes \(2\sin (x)\cos (x) - \cos (x) = 0\) or \(\cos (x) (2\sin (x) - 1) = 0\).

   From \(\cos (x) = 0\), we get \(x = \frac {\pi }{2} + \pi k\) for integers \(k\) of which only \(x = \frac {\pi }{2}\) and \(x = \frac {3\pi }{2}\) lie in \([0,2\pi )\).

   For \(2\sin (x) - 1 = 0\), we get \(\sin (x) = \frac {1}{2}\) which gives \(x = \frac {\pi }{6} + 2\pi k\) or \(x = \frac {5\pi }{6} + 2\pi k\) for integers \(k\). Of those, only \(x = \frac {\pi }{6}\) and \(x = \frac {5\pi }{6}\) lie in \([0,2\pi )\).

   Choosing test values, we get: for \(x =0\) we find \(f(0) = -1\); when \(x = \frac {\pi }{4}\) we get \(f\left (\frac {\pi }{4}\right ) =1 - \frac {\sqrt {2}}{2} = \frac {2 - \sqrt {2}}{2}\); for \(x = \frac {3\pi }{4}\) we get \(f\left (\frac {3\pi }{4}\right ) =-1 + \frac {\sqrt {2}}{2} = \frac {\sqrt {2} - 2}{2}\); when \(x=\pi \) we have \(f(\pi ) = 1\), and lastly, for \(x = \frac {7\pi }{4}\) we get \(f\left (\frac {7\pi }{4}\right ) = -1 - \frac {\sqrt {2}}{2} = \frac {-2 - \sqrt {2}}{2}\). This gives us the sign diagram below.

   

   We see \(f(x) > 0\) on \(\left (\frac {\pi }{6}, \frac {\pi }{2}\right ) \cup \left (\frac {5\pi }{6}, \frac {3\pi }{2}\right )\), so this is our answer. Geometrically, we see the graph of \(y = \sin (2x)\) is indeed above the graph of \(y = \cos (x)\) on those intervals.

3. Proceeding as above, we rewrite \(\tan (x) \geq 3\) as \(\tan (x) - 3 \geq 0\) and let \(f(x) = \tan (x) - 3\).

   We note that on \([0,2\pi )\), \(f\) is undefined at \(x =\frac {\pi }{2}\) and \(\frac {3\pi }{2}\), so those values will need the usual disclaimer on the sign diagram. ^[11] (See page ?? for a discussion of the non-standard character known as the interrobang.)

   Moving along to zeros, solving \(f(x) = \tan (x) - 3 = 0\) requires the arctangent function. We find \(x = \arctan (3) + \pi k\) for integers \(k\) and of these, only \(x = \arctan (3)\) and \(x = \arctan (3) + \pi \) lie in \([0,2\pi )\). Since \(3 > 0\), we know \(0 < \arctan (3) < \frac {\pi }{2}\) which allows us to position these zeros correctly on the sign diagram.

   To choose test values, we begin with \(x=0\) and find \(f(0) = -3\). Finding a convenient test value in the interval \(\left (\arctan (3), \frac {\pi }{2}\right )\) is a bit more challenging. Since the arctangent function is increasing and is bounded above by \(\frac {\pi }{2}\), the number \(x = \arctan (117)\) is guaranteed ^[12] (We could have chosen any value \(\arctan (t)\) where \(t > 3\).) to lie between \(\arctan (3)\) and \(\frac {\pi }{2}\). We see that \(f(\arctan (117)) = \tan (\arctan (117)) - 3 = 114\).

   For our next test value, we take \(x = \pi \) and find \(f(\pi ) = -3\), which brings us to finding a test value in the interval \(\left (\arctan (3) + \pi , \frac {3\pi }{2} \right )\).

   From \(\arctan (3) < \arctan (117) < \frac {\pi }{2}\) we get \(\arctan (3) + \pi < \arctan (117) + \pi < \frac {3\pi }{2}\) by adding \(\pi \) through the inequality. We find \(f(\arctan (117)+\pi ) = \tan (\arctan (117) + \pi ) -3 = \tan (\arctan (117)) - 3 = 114\).

   For our last test value, we choose \(x = \frac {7\pi }{4}\) and find \(f\left (\frac {7\pi }{4}\right ) = -4\). We arrive at our sign diagram below.

   

   Since we want \(f(x) \geq 0\), we see that our answer is \(\left [ \arctan (3), \frac {\pi }{2}\right ) \cup \left [\arctan (3)+\pi , \frac {3\pi }{2}\right )\). Using the graphs of \(y = \tan (x)\) and \(y = 3\), we see when the graph of the former is above (or meets) the graph of the latter. (Note, \(\arctan (3) \approx 1.249\) and \(\arctan (3) + \pi \approx 4.391\).)

1. To find the domain of \(f(x) = \csc \left (2x + \frac {\pi }{3}\right )\), we rewrite \(f\) in terms of sine as \(f(x) = \frac {1}{\sin \left (2x + \frac {\pi }{3}\right )}\). Since the sine function is defined everywhere, our only concern comes from zeros in the denominator.

   Solving \(\sin \left (2x + \frac {\pi }{3}\right ) = 0\), we get \(x = -\frac {\pi }{6} + \frac {\pi }{2} k\) for integers \(k\). In set-builder notation, our domain is \(\left \{ x \, | \, x \neq -\frac {\pi }{6} + \frac {\pi }{2} k \, \text {for integers $k$} \right \}\). To help visualize the domain, we follow the old mantra ‘When in doubt, write it out!’ We get \(\left \{ x \, | \, x \neq -\frac {\pi }{6}, \frac {2\pi }{6}, -\frac {4\pi }{6}, \frac {5\pi }{6}, -\frac {7\pi }{6}, \frac {8\pi }{6}, \ldots \right \}\), where we have kept the denominators \(6\) throughout to help see the pattern. Graphing the situation on a number line, we have

   

   Proceeding as in Section extendedinterval, we let \(x_k\) denote the \(k\)th number excluded from the domain and we have \(x_k = -\frac {\pi }{6} + \frac {\pi }{2} k = \frac {(3k-1)\pi }{6}\) for integers \(k\). The intervals which comprise the domain are of the form \(\left (x_k, x_{k+1} \right ) = \left (\frac {(3k-1)\pi }{6}, \frac {(3k+2)\pi }{6} \right )\) as \(k\) runs through the integers. Using extended interval notation, we have that the domain is

   \[ \bigcup _{k = -\infty }^{\infty } \left (\dfrac {(3k-1)\pi }{6}, \dfrac {(3k+2)\pi }{6} \right )\]

   We can check our answer by substituting in values of \(k\) to see that it matches our diagram.

2. Since the domains of \(\sin (t)\) and \(\cos (t)\) are all real numbers, the only concern when finding the domain of \(f(t) = \frac {\sin (t)}{2\cos (t) - 1}\) is division by zero so we set the denominator equal to zero and solve.

   From \(2\cos (t) - 1 = 0\) we get \(\cos (t) = \frac {1}{2}\) so \(t = \frac {\pi }{3} + 2\pi k\) or \(t = \frac {5\pi }{3} + 2\pi k\) for integers \(k\). Using set-builder notation, the domain is \(\left \{ t \, | \, t \neq \frac {\pi }{3} + 2\pi k \, \text {and} \, t \neq \frac {5\pi }{3} + 2\pi k \, \text {for integers $k$} \right \}\). Writing this out, we find the domain is \(\left \{ t \, | \, t \neq \pm \frac {\pi }{3}, \pm \frac {5\pi }{3}, \pm \frac {7\pi }{3}, \pm \frac {11\pi }{3}, \ldots \right \}\), so we have

   

   Unlike the previous example, we have two different families of points to consider, and we present two ways of dealing with this kind of situation. One way is to generalize what we did in the previous example and use the formulas we found in our domain work to describe the intervals.

   To that end, we let \(a_k = \frac {\pi }{3} + 2\pi k = \frac {(6k+1)\pi }{3}\) and \(b_k = \frac {5\pi }{3} + 2\pi k = \frac {(6k+5) \pi }{3}\) for integers \(k\). The goal now is to write the domain in terms of the \(a\)’s an \(b\)’s. We find \(a_0 = \frac {\pi }{3}\), \(a_1 = \frac {7\pi }{3}\), \(a_{-1} = -\frac {5\pi }{3}\), \(a_2 = \frac {13\pi }{3}\), \(a_{-2} = -\frac {11\pi }{3}\), \(b_0 = \frac {5\pi }{3}\), \(b_1 = \frac {11\pi }{3}\), \(b_{-1} = -\frac {\pi }{3}\), \(b_2 = \frac {17\pi }{3}\) and \(b_{-2} = -\frac {7\pi }{3}\).

   Hence, in terms of the \(a\)’s and \(b\)’s, our domain is

   \[\ldots \left (a_{-2}, b_{-2} \right ) \cup \left (b_{-2}, a_{-1} \right )\cup \left (a_{-1}, b_{-1} \right )\cup \left (b_{-1}, a_0 \right )\cup \left (a_0, b_0 \right )\cup \left (b_0, a_1 \right )\cup \left (a_1, b_1 \right )\cup \dots \]

   If we group these intervals in pairs, \( \left (a_{-2}, b_{-2} \right ) \cup \left (b_{-2}, a_{-1} \right )\), \(\left (a_{-1}, b_{-1} \right )\cup \left (b_{-1}, a_0 \right )\), \(\left (a_0, b_0 \right )\cup \left (b_0, a_1 \right )\) and so forth, we see a pattern emerge of the form \(\left (a_k, b_k \right )\cup \left (b_k, a_{k+1} \right )\) for integers \(k\) so that our domain can be written as

   \[ \bigcup _{k = -\infty }^{\infty } \left (a_k, b_k \right )\cup \left (b_k, a_{k+1} \right ) = \bigcup _{k = -\infty }^{\infty } \left (\frac {(6k+1)\pi }{3}, \frac {(6k+5) \pi }{3} \right )\cup \left (\frac {(6k+5) \pi }{3}, \frac {(6k+7)\pi }{3} \right ) \]

   A second approach to the problem exploits the periodic nature of \(f\). Since \(\cos (t)\) and \(\sin (t)\) have period \(2\pi \), it’s not too difficult to show the function \(f\) repeats itself every \(2\pi \) units. ^[14] (This doesn’t necessarily mean the period of \(f\) is \(2\pi \). The tangent function is comprised of sine and cosine, but its period is half theirs. The reader is invited to investigate the period of \(f\).) This means if we can find a formula for the domain on an interval of length \(2\pi \), we can express the entire domain by translating our answer left and right on the \(t\)-axis by adding integer multiples of \(2\pi \).

   One such interval that arises naturally from our domain work is \(\left [\frac {\pi }{3}, \frac {7\pi }{3}\right ]\). The portion of the domain here is \(\left (\frac {\pi }{3}, \frac {5\pi }{3}\right ) \cup \left (\frac {5\pi }{3}, \frac {7\pi }{3}\right )\). Adding integer multiples of \(2\pi \), we obtain the family of intervals: \(\left (\frac {\pi }{3} + 2\pi k, \frac {5\pi }{3} + 2\pi k \right ) \cup \left (\frac {5\pi }{3} + 2\pi k, \frac {7\pi }{3} + 2\pi k\right )\) for integers \(k\). We leave it to the reader to show that getting common denominators leads to our previous answer.

3. To find the domain of \(f(x) = \sqrt {1-\cot (x)}\), we first note that, due to the presence of the \(\cot (x)\) term, \(x \neq \pi k\) for integers \(k\).

   Next, we recall that for the square root to be defined, we need \(1 - \cot (x) \geq 0\). Unlike the inequalities we solved in Example TrigIneqEx1, we are not restricted here to a given interval. Our strategy is to solve this inequality over \((0,\pi )\) (the same interval which generates a fundamental cycle of cotangent) and then add integer multiples of the period, in this case, \(\pi \).

   We let \(g(x) = 1 - \cot (x)\) and set about making a sign diagram for \(g\) over the interval \((0,\pi )\) to find where \(g(x) \geq 0\). We note that \(g\) is undefined for \(x = \pi k\) for integers \(k\), in particular, at the endpoints of our interval \(x = 0\) and \(x = \pi \).

   Next, we look for the zeros of \(g\). Solving \(g(x) = 0\), we get \(\cot (x) = 1\) or \(x = \frac {\pi }{4} + \pi k\) for integers \(k\) and only one of these, \(x = \frac {\pi }{4}\), lies in \((0,\pi )\). Choosing the test values \(x = \frac {\pi }{6}\) and \(x = \frac {\pi }{2}\), we get \(g\left (\frac {\pi }{6}\right ) = 1 - \sqrt {3}\), and \(g\left (\frac {\pi }{2}\right ) = 1\). We construct the sign diagram for \(g\) over the interval \((0, \pi )\) below:

   

   We find \(g(x) \geq 0\) on \(\left [\frac {\pi }{4}, \pi \right )\). Adding multiples of the period we get our solution to consist of the intervals \(\left [\frac {\pi }{4} + \pi k, \pi + \pi k \right ) = \left [\frac {(4k+1)\pi }{4}, (k+1)\pi \right )\).

   Using extended interval notation, we have our final answer:

   \[\bigcup _{k = -\infty }^{\infty } \left [\dfrac {(4k+1)\pi }{4}, (k+1)\pi \right )\]

1. To solve \(\arcsin (2x) = \frac {\pi }{3}\), we first note that \(\frac {\pi }{3}\) is in the range of the arcsine function (so a solution exists!) Next, we exploit the inverse property of sine and arcsine from Theorem arccosinesinefunctionprops

   \[ \begin{array}{rclr} \arcsin (2x) & = & \frac {\pi }{3} & \\ \sin \left (\arcsin (2x)\right ) & = & \sin \left (\frac {\pi }{3}\right ) & \\ 2x & = & \frac {\sqrt {3}}{2} & \text {Since $\sin (\arcsin (u)) = u$} \\ x & = & \frac {\sqrt {3}}{4} & \\ \end{array} \]

   Below we see the graphs of \(y = \arcsin (2x)\) and \(y = \frac {\pi }{3}\), intersect at \(x = \frac {\sqrt {3}}{4} \approx 0.4430\).

2. Our first step in solving \(4\arccos (t)-3\pi = 0\) is to isolate the arccosine. We get \(\arccos (t) = \frac {3\pi }{4}\). Since \(\frac {3\pi }{4}\) is in the range of arccosine, we may apply Theorem arccosinesinefunctionprops

   \[ \begin{array}{rclr} \arccos (t) & = & \frac {3\pi }{4} & \\ \cos \left (\arccos (t)\right ) & = & \cos \left (\frac {3\pi }{4}\right ) & \\ t & = & -\frac {\sqrt {2}}{2} & \text {Since $\cos (\arccos (u)) = u$} \\ \end{array} \]

   Below we see the graph of \(y = 4\arccos (t) - 3\pi \) crosses \(y = 0\) at \(t=-\frac {\sqrt {2}}{2} \approx -0.7071\).

3. From \(3 \, \text {arcsec}(2x-1) + \pi = 2 \pi \), we get \(\text {arcsec}(2x-1) = \frac {\pi }{3}\). Regardless of how the range of arcsecant is chosen, since \(0 \leq \frac {\pi }{3} < \frac {\pi }{2}\), both Theorems arcsecantcosecantfunctionprops1 arcsecantcosecantfunctionprops2, apply:

   \[ \begin{array}{rclr} \text {arcsec}(2x-1)& = & \frac {\pi }{3} & \\ \sec (\text {arcsec}(2x-1)) & = & \sec \left (\frac {\pi }{3}\right ) & \\ 2x -1 & = & 2 & \text {Since $\sec (\text {arcsec}(u)) = u$} \\ x & = & \frac {3}{2} \\ \end{array} \]

   Below we see the graphs of \(y=3 \, \text {arcsec}(2x-1) + \pi \) and \(y = 2\pi \) intersect at \(x=\frac {3}{2} = 1.5\).

4. With the presence of both \(\arctan ^{2}(t)\) ( \(= (\arctan (t))^2\)) and \(\arctan (t)\), we substitute \(u = \arctan (t)\) to reveal a quadratic in disguise: \(4u^2 -3\pi u - \pi ^2 = 0\).

   Factoring, (don’t let the \(\pi \) throw you!) we get \((4u+\pi )(u - \pi ) = 0\), so \(u = \arctan (t) = -\frac {\pi }{4}\) or \(u = \arctan (t) = \pi \).

   Since \(-\frac {\pi }{4}\) is in the range of arctangent, but \(\pi \) is not, we only get solutions from the first equation. Using Theorem arctangentcotangentfunctionprops, we get

   \[ \begin{array}{rclr} \arctan (t) & = & -\frac {\pi }{4} & \\ \tan (\arctan (t)) & = & \tan \left (-\frac {\pi }{4}\right ) & \\ t & = & -1 & \text {Since $\tan (\arctan (u)) = u$.} \\ \end{array}\]

   We verify this result graphically below.

5. Since the inverse circular functions are continuous on their domains, we can solve inequalities featuring these functions using sign diagrams.

   Since all of the nonzero terms of \(\pi ^2-4\arccos ^{2}(x) < 0\) are on one side of the inequality, we let \(f(x) = \pi ^2-4\arccos ^{2}(x)\) and note the domain of \(f\) is limited by the \(\arccos (x)\) to \([-1,1]\).

   Next, we find the zeros of \(f\) by setting \(f(x) = \pi ^2-4\arccos ^{2}(x) = 0\). We get \(\arccos (x) = \pm \frac {\pi }{2}\), and since the range of arccosine is \([0,\pi ]\), we focus our attention on \(\arccos (x) = \frac {\pi }{2}\).

   Using Theorem arccosinesinefunctionprops, we get \(x = \cos \left (\frac {\pi }{2}\right ) = 0\) as our only zero which breaks our domain \([-1,1]\) into two test intervals: \([-1,0)\) and \((0,1]\). Choosing test values \(x = \pm 1\), we get \(f(-1) = -3\pi ^2 < 0\) and \(f(1) = \pi ^2 > 0\). Our sign diagram is below.

   

   Since we are looking for where \(f(x) = \pi ^2-4\arccos ^{2}(x) < 0\), our answer is \([-1,0)\).

   Geometrically, we find the graph of \(y = \pi ^2-4\arccos ^{2}(x)\) is below \(y = 0\) (the \(x\)-axis) on \([-1,0)\).

6. As in the previous problem, we will use a sign diagram to solve \(4 \, \text {arccot}(3t) > \pi \). Our first step is to rewrite the inequality as \(4 \, \text {arccot}(3t) - \pi > 0\).

   We let \(f(t) = 4 \, \text {arccot}(3t) - \pi \), and find the domain of \(f\) is all real numbers, \((-\infty , \infty )\).

   To find the zeros of \(f\), we set \(f(t) = 4 \, \text {arccot}(3t) - \pi = 0\) and solve. We get \(\text {arccot}(3t) = \frac {\pi }{4}\), and since \(\frac {\pi }{4}\) is in the range of arccotangent, we may apply Theorem arctangentcotangentfunctionprops and solve

   \[ \begin{array}{rclr} \text {arccot}(3t) & = & \frac {\pi }{4} & \\ \cot (\text {arccot}(3t)) & = & \cot \left (\frac {\pi }{4}\right ) & \\ 3t & = & 1 & \text {Since $\cot (\text {arccot}(u)) = u$.} \\ t & = & \frac {1}{3} & \\ \end{array}\]

   Next, we make a sign diagram for \(f\). Since the domain of \(f\) is all real numbers, and there is only one zero of \(f\), \(t = \frac {1}{3}\), we have two test intervals, \(\left (-\infty , \frac {1}{3}\right )\) and \(\left (\frac {1}{3}, \infty \right )\).

   Ideally, we wish to find test values \(t\) in these intervals so that \(\text {arccot}(3t)\) corresponds to one of our oft-used ‘common’ angles. After a bit of computation, ^[15] (Set \(3t\) equal to the cotangents of the ‘common angles’ and choose accordingly.) we choose \(t=0\) for the interval \(\left (-\infty , \frac {1}{3}\right )\) and \(t = \frac {\sqrt {3}}{3}\) for the interval \(\left (\frac {1}{3}, \infty \right )\). We find \(f(0) = \pi > 0\) and \(f\left (\frac {\sqrt {3}}{3}\right ) = -\frac {\pi }{3} < 0\). Our sign diagram is below.

   

   Since we are looking for where \(f(t) > 0\), we get our answer \(\left (-\infty , \frac {1}{3}\right )\). Graphically, we see the graph of \(y = 4 \, \text {arccot}(3t)\) is above the horizontal line \(y = \pi \) on \(\left (-\infty , \frac {1}{3}\right )= \left (-\infty , 0.\overline {3} \right )\).

To find \(k\), we use Hooke’s Law \(F = kd\). We know the object weighs \(64\) lbs. and stretches the spring \(8\) ft.. Using \(F = 64\) and \(d = 8\), we get \(64 = k \cdot 8 \), or \(k = 8 \frac {\text {lbs.}}{\text {ft.}}\).

To find \(m\), we use \(w = mg\) with \(w = 64\) lbs. and \(g =32 \frac {\text {ft.}}{s^2}\). We get \(m = 2\) slugs. We can now proceed to apply Theorem 1.

1. With \(k = 8\) and \(m = 2\), we get \(\omega = \sqrt {\frac {k}{m}} = \sqrt {\frac {8}{2}} = 2\). Since the object is released 3 feet below the equilibrium position ‘from rest,’ \(x_0 = 3\) and \(v_0 = 0\). Therefore, \(A = \sqrt {x_0^2 + \left ( \frac {v_0}{\omega }\right )^2} = \sqrt {3^2 + 0^2} = 3\).

   To determine the phase \(\phi \), we have \(A\sin (\phi ) = x_0\), which in this case gives \(3 \sin (\phi ) = 3\) so \(\sin (\phi ) = 1\). Only \(\phi = \frac {\pi }{2}\) and angles coterminal to it satisfy this condition, so we pick ^[22] (For confirmation, we note that \(A\omega \cos (\phi ) = v_0\), which in this case reduces to \(6\cos (\phi ) = 0\).) the phase to be \(\phi = \frac {\pi }{2}\). Hence, the equation of motion is \(x(t) = 3\sin \left (2t + \frac {\pi }{2}\right )\).

   To find when the object passes through the equilibrium position we solve \(x(t)= 3\sin \left (2t + \frac {\pi }{2}\right ) = 0\). Going through the usual analysis we find \(t = -\frac {\pi }{4} + \frac {\pi }{2} k\) for integers \(k\). Since we are interested in the first time the object passes through the equilibrium position, we look for the smallest positive \(t\) value which in this case is \(t = \frac {\pi }{4} \approx 0.78\) seconds after the start of the motion.

   Common sense suggests that if we release the object below the equilibrium position, the object should be traveling upwards when it first passes through it. To check this answer, we graph one cycle of \(x(t)\). Since our applied domain in this situation is \(t \geq 0\), and the period of \(x(t)\) is \(T = \frac {2\pi }{\omega } = \frac {2\pi }{2} = \pi \), we graph \(x(t)\) over the interval \([0,\pi ]\). Remembering that \(x(t) > 0\) means the object is below the equilibrium position and \(x(t) < 0\) means the object is above the equilibrium position, the fact our graph is crossing through the \(t\)-axis from positive \(x\) to negative \(x\) at \(t = \frac {\pi }{4}\) confirms our answer.

2. The only difference between this problem and the previous problem is that we now release the object with an upward velocity of \(8 \, \frac {\text {ft}}{s}\). We still have \(\omega = 2\) and \(x_0 = 3\), but now we have \(v_0 = -8\), the negative indicating the velocity is directed upwards.

   Here, we get \(A = \sqrt {x_0^2 + \left ( \frac {v_0}{\omega }\right )^2} = \sqrt {3^2 + (-4)^2} = 5\). From \(A\sin (\phi ) = x_0\), we get \(5\sin (\phi ) = 3\) which gives \(\sin (\phi ) = \frac {3}{5}\). From \(A\omega \cos (\phi ) = v_0\), we get \(10\cos (\phi ) = -8\), or \(\cos (\phi ) = -\frac {4}{5}\).

   Hence, \(\phi \) is a Quadrant II angle which we can describe in terms of either arcsine or arccosine. Since the range of arccosine covers Quadrant II, we choose to express \(\phi \) in terms of the arccosine: \(\phi = \arccos \left ( -\frac {4}{5} \right )\). Hence, \(x(t)= 5 \sin \left (2t + \arccos \left (-\frac {4}{5}\right ) \right )\).

   Since the amplitude of \(x(t)\) is \(5\), the object will travel at most \(5\) feet above the equilibrium position. To find when this happens, we solve the equation \(x(t)= 5 \sin \left (2t +\arccos \left ( -\frac {4}{5} \right ) \right )= -5\), the negative once again signifying that the object is above the equilibrium position.

   Going through the usual machinations, we get \(t = -\frac {1}{2} \arccos \left ( -\frac {4}{5} \right )-\frac {\pi }{4} + \pi k\) for integers \(k\). The smallest (positive) of these values occurs when \(k=1\), that is, \(t =-\frac {1}{2} \arccos \left ( -\frac {4}{5} \right )+ \frac {3\pi }{4} \approx 1.107\) seconds after the start of the motion.

   Graphing \(x(t) = 5 \sin \left (2t + \arccos \left ( -\frac {4}{5} \right ) \right )\), we find the coordinates of the first relative minimum of to be approximately \((1.107,-5)\) .